6
$\begingroup$

I know that two data sets can have the same Kendall's $\tau$ but different Pearson's $\rho$.

  1. What about the opposite? Can two different data sets have the same Pearson's $\rho$, but different Kendall's $\tau$? Or rather, if we know the correlation matrix Pearson's $\rho_{ij}$, does it correspond to precisely one matrix of Kendall's $\tau_{ij}$, obtained by transforming the original matrix to Kendall's $\tau$ by $$\frac{2}{\pi}\arcsin{(\rho_{ij})}$$
  2. If we sample from a multivariate distribution with a certain population correlation matrix $\rho_{ij}$, is this equivalent to sampling from a distribution with a population matrix $\tau_{ij}$ obtained by the previously mentioned transformation? I.e. do the samples, which have a sample Pearson correlation matrix $\Sigma$ and associated Kendall's correlation matrix, come from the population with this $\tau_{ij}$ obtained by transforming population $\rho_{ij}$?
$\endgroup$

2 Answers 2

12
$\begingroup$

Can two different data sets have the same Pearson's ρ, but different Kendall's τ?

Anscombe's quartet gives you four two-dimensional datasets with (almost) identical Pearson correlations, but their Kendall correlations are quite different. In R:

> with(anscombe,cor(x1,y1,method="pearson"))
[1] 0.8164205
> with(anscombe,cor(x2,y2,method="pearson"))
[1] 0.8162365
> with(anscombe,cor(x1,y1,method="kendall"))
[1] 0.6363636
> with(anscombe,cor(x2,y2,method="kendall"))
[1] 0.5636364

In general, there is no bijective transformation between the two correlations. You can also see this by the fact that small changes in data points will always change the Pearson correlation continuously - but the Kendall correlation will not change at first and suddenly change when data points change position, because the Kendall correlation only looks at whether two points are to the left or the right, respectively above or below each other.

$\endgroup$
6
  • $\begingroup$ Thanks! What about question (2)? Is there a bijective transformation between the two in population? I found the formula above, which has been proven, but obviously does not hold for finite samples. But does it hold for population values? $\endgroup$
    – Novic
    Mar 31, 2023 at 9:38
  • $\begingroup$ Per my bottom paragraph, there should not be any bijection between the two correlation coefficients at all, so I am puzzled how you came by that formula. Can you give us the source? $\endgroup$ Mar 31, 2023 at 11:33
  • $\begingroup$ Yes, it's Kendall's formula, e.g. used in this article: digitalcommons.wayne.edu/cgi/… (p.2, assumed normal distribution and a relatively large sample?). It's also used to compute theoretical $\tau$ for parameters (rho) of a Gaussian copula, e.g. tnagler.github.io/VineCopula/reference/BiCopPar2Tau.html $\endgroup$
    – Novic
    Mar 31, 2023 at 11:46
  • $\begingroup$ this question provides further sources on where it's proven (the formula I've written above is apparently Greiner's Equality) stats.stackexchange.com/questions/133460/… $\endgroup$
    – Novic
    Mar 31, 2023 at 12:08
  • 1
    $\begingroup$ Ah, but those resources are under the additional assumption that we are working in a Gaussian copula, that is a very restrictive assumption, and your post was about multivariate distributions in general. If you want to ask about the very specific case of a Gaussian copula, I would recommend you post a new question. (Ideally, link the two questions together.) $\endgroup$ Mar 31, 2023 at 13:45
4
$\begingroup$

An indirect approach:

If two $\rho_1 > \rho_2$ can relate to the same $\tau$, then imagine adjusting the dataset a little such that we increase all these three values.

We should be able to arrive to some values $\rho^\prime_1 > \rho_1$, $\rho^\prime_2 > \rho_2$ and $\tau^\prime > \tau$. This should be possible in a continuous way such that we get at some point $\rho^\prime_2 = \rho_1$.

That means that we have got the same Pearson's $\rho$, but with different $\tau^\prime$ and $\tau$.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks! Intuitively I understand why this should be possible, but could you maybe spell out the exact math behind it? $\endgroup$
    – Novic
    Mar 31, 2023 at 9:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.